A box contains 15 tickets numbered 1, 2, .... | vedclass

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Question

(c) On trial, $n = 15$ since any of the $15$ numbers can be on the selected coin and $m = 9$ since the largest number is $9$ and so it can be $1$ or $

Vedclass · Accepted Answer

${\left( {\frac{3}{5}} \right)^7}$

Vedclass · Answer

(c) On trial, $n = 15$ since any of the $15$ numbers can be on the selected coin and $m = 9$ since the largest number is $9$ and so it can be $1$ or $2$ or $3.........$ or $9$.
We have required probability $ = {\left( {\frac{9}{{15}}} \right)^7} = {\left( {\frac{3}{5}} \right)^7}.$

A box contains $15$ tickets numbered $1, 2, ....... 15$. Seven tickets are drawn at random one after the other with replacement. The probability that the greatest number on a drawn ticket is $9$, is

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